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# Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra based on Recursion from AMC 10 A, 2019. You may use sequential hints to solve the problem.

## Recursion- AMC-10A, 2019- Problem 15

A sequence of numbers is defined recursively by $a_{1}=1, a_{2}=\frac{3}{7},$ and $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

for all $n \geq 3$ Then $a_{2019}$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q ?$

• $2020$
• $4039$
• $6057$
• $6061$
• $8078$

### Key Concepts

Algebra

Recursive formula

## Suggested Book | Source | Answer

Pre College Mathematics

#### Source of the problem

AMC-10A, 2019 Problem-15

#### Check the answer here, but try the problem first

$8078$

## Try with Hints

#### First Hint

The given expression is $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$ and given that $a_{1}=1, a_{2}=\frac{3}{7}$. we have to find out $$a_{2019}$$?

at first we may use recursive formula we can find out $$a_3$$ , $$a_4$$ with the help of $$a_1$$, $$a_2$$. later we can find out $$a_n$$

Now can you finish the problem?

#### Second Hint

Given that $a_{n}=\frac{a_{n-2} \cdot a_{n-1}}{2 a_{n-2}-a_{n-1}}$

Now $$n=3$$ then $a_{3}=\frac{a_{(3-2)} \cdot a_{(3-1)}}{2 a_{(3-2)}-a_{(3-1)}}$

$$\Rightarrow$$ $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

$$\Rightarrow$$ $a_{3}=\frac{a_{(1)} \cdot a_{(2)}}{2 a_{(1)}-a_{(2)}}$

$$\Rightarrow$$ $a_{3}=\frac{1*\frac{3}{7}}{2*1-\frac{3}{7}}$

$$\Rightarrow$$ $a_{3}=\frac{3}{7}$

Similarly if we put $$n=4$$ we get $$a_4=\frac{3}{15}$$ (where $a_{1}=1, a_{2}=\frac{3}{7}$,$$a_3=\frac{3}{7}$$)

Continue this way we $a_{n}=\frac{3}{4 n-1}$

So can you find out the value of $$a_{2019}$$?

Now Can you finish the Problem?

#### Third Hint

Now $a_{n}=\frac{3}{4 n-1}$

Put $$n=2019$$

$a_{2019}=\frac{3}{8075}$ which is the form of $$\frac{p}{q}$$

Therefore $$p+q=8078$$